An urn contains 120 balls numbered 1 to 120. Find the probability of choosing 3 balls without replacement that includes one ball numbered “11”.
My attempt: Picking "11" would have the probability 1/120
The other two numbers would have probability of 119/119 and 118/118 since my sample space is decreasing but it can be any number.
So I would just be left with 1/120.
But then what if I pick two other balls and then the "11"? Would that not have the probability of 1/118?
On
The probability of picking one ball that is numbered 11 is:
Probability of picking the ball 11 first + Probability of picking the ball 11 second + Probability of picking the ball 11 third.
Which is: $$\frac{1}{120}*\frac{119}{119}*\frac{118}{118}+\frac{119}{120}*\frac{1}{119}*\frac{118}{118}+\frac{119}{120}*\frac{118}{119}*\frac{1}{118}$$
Which equals $\frac{1}{40}$ or 2.5% chance
Since order doesn’t matter, you’re better off counting: there are $\binom{119}2=\frac{119\cdot118}2=119\cdot59$ pairs of balls that do not include the $11$-ball, so there are $119\cdot59$ sets of $3$ balls that include the $11$-ball. There are $\binom{120}3=\frac{120\cdot119\cdot118}{3\cdot2\cdot1}=40\cdot119\cdot59$ sets of $3$ balls altogether. Thus, the desired probability is
$$\frac{119\cdot59}{40\cdot119\cdot59}=\frac1{40}\;.$$
If you approach the problem your way, you have to consider three cases. If you draw the balls one at a time, your $\frac1{120}$ is the probability of drawing the $11$-ball first and then two others. The probability of drawing it second is
$$\frac{119}{120}\cdot\frac1{119}\cdot\frac{118}{118}=\frac1{120}\;,$$
and the probability of drawing it third is
$$\frac{119}{120}\cdot\frac{118}{119}\cdot\frac1{118}=\frac1{120}\;.$$
Add the three, and you get the correct value of $\frac1{40}$.